Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is

Options

A. 350
B. 500
C. 200
D. 300

Explanation

To solve the problem of forming teams of 2 girls and 3 boys from a class of 5 girls and 7 boys, where two specific boys (A and B) refuse to be on the same team, we can follow these steps:

### Step 1: Calculate the total number of ways to form a team without restrictions.
We need to select 2 girls from 5 and 3 boys from 7. The number of ways to choose 2 girls from 5 is given by the combination formula \( \binom{n}{r} \), which represents the number of ways to choose \( r \) items from \( n \) items without regard to the order of selection.

\[ \text{Total ways} = \binom{5}{2} \times \binom{7}{3} \]

Calculating each part:
- \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
- \( \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)

Thus, the total number of ways is:
\[ \text{Total ways} = 10 \times 35 = 350 \]

### Step 2: Calculate the number of ways in which boys A and B are on the same team.
If boys A and B are on the same team, we can treat them as a single unit. This means we have already chosen 2 boys (A and B), and we need to select 1 more boy from the remaining 5 boys.

The number of ways to choose 1 boy from the remaining 5 is:
\[ \binom{5}{1} = 5 \]

We still need to select 2 girls from the 5 girls:
\[ \binom{5}{2} = 10 \]

Thus, the number of ways with boys A and B on the same team is:
\[ \text{Ways with A and B together} = 10 \times 5 = 50 \]

### Step 3: Calculate the number of ways in which boys A and B are not on the same team.
To find the number of teams where boys A and B are not together, we subtract the number of teams where they are together from the total number of teams:

\[ \text{Ways with A and B not together} = \text{Total ways} - \text{Ways with A and B together} \]
\[ \text{Ways with A and B not together} = 350 - 50 = 300 \]

### Final Answer:
The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, ensuring that boys A and B are not on the same team, is **300**.